NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers

Class 8 exponents and powers NCERT solutions - Topic 12.2 Powers With Negative Exponents

Question:(i) Find the multiplicative inverse of the following.

$2^{-4}$

Answer:

The detailed explanation for the question is written below,

The multiplicative inverse is $\frac{1}{a^m}$

So, the multiplicative inverse of $2^{-4}$ is $2^4$

Question:(ii) Find the multiplicative inverse of the following.

${10}^{-5}$

Answer:

Here is the detailed solution for the above question,

As we know,

The multiplicative inverse of $a^m$ is $\frac{1}{a^m}$

So, the multiplicative inverse of ${10}^{-5}$ is ${10}^5$

Question:(iv) Find the multiplicative inverse of the following.

$5^{-3}$

Answer:

we know,

The multiplicative inverse of $a^m$ is $\frac{1}{a^m}$

So, for $5^{-3}$ multiplicative inverse is $5^3$

Question:(v) Find the multiplicative inverse of the following.

${10}^{-100}$

Answer:

The multiplicative inverse of $a^m$ is $\frac{1}{a^m}$

So, the multiplicative inverse of ${10}^{-100}$ is ${10}^{100}$

NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers Topic 12.2 Powers With Negative Exponents

Question:(i) Expand the following numbers using exponents.

1025.63

Answer:

1025.63 = $1\times {10}^3+o\times {10}^2+2\times {10}^1+5\times {10}^0+6\times {10}^{-1}+3\times {10}^{-2}$

Question:(ii) Expand the following numbers using exponents.

1256.249

Answer:

1256.249

$=1\times {10}^3+2\times {10}^2+5\times {10}^1+6\times {10}^0+2\times {10}^{-1}+4\times {10}^{-2}+9\times {10}^{-3}$

NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers - Topic 12.3 Laws Of Exponents

Question:1(i) Simplify and write in exponential form.

$(-2{)}^{-3}\times (-2{)}^{-4}$

Answer:

this is simplified as follows

$(-2{)}^{-3}\times (-2{)}^{-4}$

$=\frac{1}{(-2{)}^3}\times \frac{1}{(-2{)}^4}$

$=\frac{1}{(-2{)}^{3+4}}=\frac{1}{(-2{)}^7}$

$=(-2{)}^{-7}$

Question:1(ii) Simplify and write in exponential form .

$p^3\times p^{-10}$

Answer:

this is simplified as follows

$p^3\times p^{-10}$

$p^{3-10}$ ............. $[a^m\times a^n=a^{m+n}]$

$=p^{-7}$

Question:1(iii) Simplify and write in exponential form.

$3^2\times 3^{-4}\times 3^6$

Answer:

this can be simplified as follows

$3^2\times 3^{-4}\times 3^6$

$=3^{2+(-4)+6}$ ............. $[a^m\times a^n\times a^o=a^{m+n+o}]$

$=3^4=81$

Class 8 maths chapter 12 question answer - Exercise: 12.1

Question:1 (i) Evaluate.

$3^{-2}$

Answer:

The detailed explanation for the above-written question is as follows,

We know that,

$a^{-m}=\frac{1}{a^m}$

So, here m =2 and a = 3

$3^{-2}=\frac{1}{3^2}=\frac{1}{3}\times \frac{1}{3}=\frac{1}{9}$

Question: 1(ii) Evaluate.

$(-4{)}^{-2}$

Answer:

The detailed explanation for the above-written question is as follows

We know that,

$a^{-m}=\frac{1}{a^m}$

So, here (a = -4) and (m = 2)

Then according to the law of exponent

$(-4{)}^{-2}=\frac{1}{(-4{)}^2}=\frac{1}{(-4)}\times \frac{1}{(-4)}=\frac{1}{16}$ [ negative $\times$ negative = positive]

Question: 1(iii) Evaluate.

${\left(\frac{1}{2}\right)}^{-5}$

Answer:

The detailed solution for the above-written question is as follows

We know that,

$(\frac{a}{b}{)}^m=\frac{a^m}{b^m}$ $\mathrm{\&}$ $a^{-m}=\frac{1}{a^m}$

So, here

a = 1 and b = 2 and m =-5

According to the law of exponent

$(\frac{1}{2}{)}^{-5}=\frac{1^{-5}}{2^{-5}}=\frac{-1}{2^{-5}}$

$=(-1)\times -2\times -2\times -2\times -2\times -2=32$

Question: 2(i) Simplify and express the result in power notation with a positive exponent.

$(-4{)}^5÷(-4{)}^8$

Answer:

The detailed solution for the above-written question is as follows

We know the exponential formula

$\frac{a^m}{a_n}=a^{m-n}$ and $a^{-m}=\frac{1}{a^m}$

So according to this

a = -4, m = 5 and n = 8

$\frac{4^5}{4_8}=-4^{5-8}=-4^{-3}$

$=(-\frac{1}{4})\times -(\frac{1}{4})\times (-\frac{1}{4})=-\frac{1}{64}$

Question: 2(ii) Simplify and express the result in power notation with positive exponent.

${\left(\frac{1}{2^3}\right)}^2$

Answer:

The detailed solution for the above-written question is as follows

We know the exponential formula

$\frac{a^m}{b^m}=(\frac{a}{b}{)}^m$ and $a^{-m}=\frac{1}{a^m}$ and $(a^m{)}^n=a^{mn}$

So, we have given

a = 1, b=2

By using above exponential law,

$\frac{a^m}{b^m}=\frac{1}{(2^3{)}^2}=\frac{1}{2^6}$

$=\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{64}$

Question: 2(iii) Simplify and express the result in power notation with a positive exponent.

$(-3{)}^4\times {\left(\frac{5}{3}\right)}^4$

Answer:

The detailed solution for the above-written question is as follows,

We know the exponential formula

$(\frac{a}{b}{)}^m=\frac{a^m}{b^m}$

So, $(-3{)}^4\times (\frac{5}{3}{)}^4=(-3{)}^4\times \frac{5^4}{3^4}=\frac{625}{1}$

Question: 2(iv) Simplify and express the result in power notation with a positive exponent.

$(3^{-7}÷3^{-10})\times 3^{-5}$

Answer:

The detailed explanation for the above-written question is as follows

As we know the exponential form

$\frac{a^m}{b^n}=a^{m-n}\mathrm{\&}(a^m\times a^n)=a^{m+n}$

By using these two form we get,

$\frac{3^{-7}}{3^{-10}}\times (3{)}^{-5}=3^{-7-(-10)}\times (3{)}^{-5}$

$=3^3\times (3{)}^{-5}=3^{-2}$

$=\frac{1}{3}\times \frac{1}{3}=\frac{1}{9}$

Question:2(v) Simplify and express the result in power notation with positive exponent.

$2^{-3}\times (-7{)}^{-3}$

Answer:

The detailed solution for the above-written question is as follows,

we know the exponential forms

$a^{-m}=\frac{1}{a^m}$ & $a^m\times b^m=(ab{)}^m$

So, according to our data,

here initially we use first forms and then the second one.

$2^{-3}\times (-7{)}^{-3}=\frac{1}{2^3}\times \frac{1}{(-7{)}^3}$

$=\frac{1}{(2\times -7{)}^3}=\frac{1}{(-14{)}^3}$

Question:3(i) Find the value of.

$(3^0+4^{-1})\times 2^2$

Answer:

The detailed explanation for the above-written question is as follows,

As we know that $a^0=1$

So, $3^0=1$

now,

$=(1+\frac{1}{4})\times 2^2$

$=\frac{5}{4}\times 2^2\Rightarrow \frac{5}{2^2}\times 2^2=5/1$

Question: 3(ii) Find the value of.

$(2^{-1}\times 4^{-1})÷2^{-2}$

Answer:

The detailed explanation for the above-written question is as follows

Rewrite the equation

$(2^{-1}\times 4^{-1})÷2^{-2}$ $=(2^{-1}\times 2^{-2})÷2^{-2}$

$=(2^{-1+(-2)})÷2^{-2}$ ................................. $a^m\times a^n=a^{(m+n)}$

$=(2^{-3})÷2^{-2}=2^{-3-(-2)}$ ........................ $a^m÷a^n=a^{(m-n)}$

$=2^{-1}=\frac{1}{2}$

Question: 3(iii) Find the value of.

${\left(\frac{1}{2}\right)}^{-2}+{\left(\frac{1}{3}\right)}^{-2}+{\left(\frac{1}{4}\right)}^{-2}$

Answer:

The detailed explanation for the above-written question is as follows,

This is the exponential form

$(a/b{)}^m=\frac{a^m}{b^m}$

So, ${\left(\frac{1}{2}\right)}^{-2}+{\left(\frac{1}{3}\right)}^{-2}+{\left(\frac{1}{4}\right)}^{-2}$

$=\frac{1}{2^{-2}}+\frac{1}{3^{-2}}+\frac{1}{4^{-2}}$ .......................using this form $a^m=\frac{1}{a^m}$

$=2^2+3^2+4^2$

= 4+9+16

= 29

Question: 3(iv) Find the value of.

$(3^{-1}+4^{-1}+5^{-1}{)}^0$

Answer:

since we know that

$a^0=1$

$(3^{-1}+4^{-1}+5^{-1}{)}^0$ $=1$

Question: 3(v) Find the value of.

${\left\{{\left(\frac{-2}{3}\right)}^{-2}\right\}}^2$

Answer:

The detailed explanation for the above-written question is as follows

${\left\{{\left(\frac{-2}{3}\right)}^{-2}\right\}}^2$

$=(-2/3{)}^{-2\times 2}$ .............. BY using these form of exponential $(a^m{)}^n=a^{mn}$

$(-2/3{)}^{-4}=(-3/2{)}^4$ ......... use this $a^{-m}=\frac{1}{a^m}$

$=\frac{81}{16}$

Question:4(i) Evaluate

$\frac{8^{-1}\times 5^3}{2^{-4}}$

Answer:

The detailed explanation for the above written question is as follows

$\frac{8^{-1}\times 5^3}{2^{-4}}$

after rewriting the above equation we get,

$\frac{2^{-3}\times 5^3}{2^{-4}}$ $=2^{-3-(-4)}\times 5^3$ ...........as we know that $\frac{a^m}{a^n}=a^{m-n}$

$$\begin{gathered} =2^1\times 5^3 \\ =2\times 125=250 \end{gathered}$$

An alternate method,

$=\frac{5^3}{2^{-4}\times 2^3}$

here you can use first $a^{-m}=\frac{1}{a^m}$ and after that use $a^m\times a^n=a^{m+n}$

Question: 4(ii) Evaluate

$(5^{-1}\times 2^{-1})\times 6^{-1}$

Answer:

The detailed explanation for the above-written question is as follows

We clearly see that this is in the form of $a^{-m}=\frac{1}{a^m}$

So, $(5^{-1}\times 2^{-1})\times 6^{-1}=(\frac{1}{5}\times \frac{1}{2})\times \frac{1}{6}$

$=\frac{1}{60}$

Question: 5 Find the value of $m$ for which

$5^m÷5^{-3}=5^5$

Answer:

We have,

$a^m÷a^n=a^{m-n}$

Here a = 5 and n =-3 and m-n = 5

therefore,

$5^m÷5^{-3}=5^{m+3}=5^5$

By comparing from both sides we get

m+3 = 5

m= 2

Question: 6(i) Evaluate

${\{{\left(\frac{1}{3}\right)}^{-1}-{\left(\frac{1}{4}\right)}^{-1}\}}^{-1}$

Answer:

The detailed solution for the above-written question is as follows

${\{{\left(\frac{1}{3}\right)}^{-1}-{\left(\frac{1}{4}\right)}^{-1}\}}^{-1}$

$=[(1\times 3)-(1\times 4){]}^{-1}$ .............by using $a^{-m}=\frac{1}{a^m}$

$=[3-4{]}^{-1}$

$$\begin{gathered} =[-1{]}^{-1} \\ =-1 \end{gathered}$$

Question: 6(ii) Evaluate

${\left(\frac{5}{8}\right)}^{-7}\times {\left(\frac{8}{5}\right)}^{-4}$

Answer:

The detailed solution for the above-written question is as follows

${\left(\frac{5}{8}\right)}^{-7}\times {\left(\frac{8}{5}\right)}^{-4}$

$=\frac{5^{-7}}{8^{-7}}\times \frac{8^{-4}}{5^{-4}}$ ................ using the form $\frac{a^m}{b^m}=(a/b{)}^m$

$=5^{-7+4}\times 8^{-4+7}$ ............using $a^m÷a^n=a^{m-n}$

$=5^{-3}\times 8^{+3}$

$$\begin{gathered} =\frac{8^3}{5^3} \\ =\frac{512}{125} \end{gathered}$$

Question: 7(i) Simplify

$\frac{25\times t^{-4}}{5^{-3}\times 10\times t^{-8}}(t\ne 0)$

Answer:

The detailed solution for the above-written question is as follows

$\frac{25\times t^{-4}}{5^{-3}\times 10\times t^{-8}}(t\ne 0)$

we can write $25=5^2$

So, after rewriting the equation,

$\frac{5^2\times t^{-4}}{5^{-3}\times 10\times t^{-8}}$

$=\frac{5^{2+3}\times t^{-4+8}}{10}$ .................using the form $[a^m÷a^n=a^{m-n}]$

$=\frac{5^5\times t^4}{10}$ .............(By expanding we have now)

$=\frac{625t^4}{2}$

Question: 7(ii) Simplify.

$\frac{3^{-5}\times {10}^{-5}\times 125}{5^{-7}\times 6^{-5}}$

Answer:

The detailed solution for the above-written question is

$\frac{3^{-5}\times {10}^{-5}\times 125}{5^{-7}\times 6^{-5}}$

we can write 125 = $5^3$ and $6^{-5}$ can be written as $(2\times 3{)}^{-5}$

Now, rewriting the equation, we get

$=\frac{3^{-5}\times {10}^{-5}\times 5^3}{5^{-7}\times (2\times 3{)}^{-5}}$

$=\frac{3^{-5}\times {10}^{-5}\times 5^{3+7}}{(2\times 3{)}^{-5}}$ .............by using $[a^m÷a^n=a^{m-n}]$

$=\frac{{10}^{-5}\times 5^{10}}{(2{)}^{-5}}$ .....................Use $[a^m÷a^n=a^{m-n}]$

$5^{10-5}=5^5=3125$ .........................As $[{10}^{-5}=(2\times 5{)}^{-5}=2^{-5}\times 5^{-5}]$ . $2^{-5}$ can be cancelled out with the denominator $2^{-5}$

Class 8 maths chapter 12 NCERT solutions - Topic 12.4 Use Of Exponents To Express Small Numbers In Standard Form

Question: 1(i) Write the following numbers in standard form.

0.000000564

Answer:

the standard form of 0.000000564 is

$\frac{564}{1000000000}=5.64\times {10}^{-7}$

Question: 1(ii) Write the following numbers in standard form.

0.0000021

Answer:

The standard form 0.0000021 is

$=\frac{21}{10000000}=2.1\times {10}^{-6}$

Question: 1(iii) Write the following numbers in standard form.

21600000

Answer:

The standard form 21600000 is

$=2.16\times {10}^7$

Question: 1(iv) Write the following numbers in standard form.

15240000

Answer:

The standard form 15240000

$=1.524\times {10}^7$

Question: 2 Write all the facts given in the standard form .

Answer:

1. Distance between sun and earth $1.496\times {10}^{11}m$

2. speed of light is $3\times {10}^{8}m/s$

3. The avg. diameter of red blood cells is $(7\times {10}^{-6}mm)$

4. the distance of the moon from the earth is $(3.84467\times {10}^{8}m)$

5. size of the plant cell is $(1.275\times {10}^{-5}m)$

6. The diameter of the wire on a computer chip is $(3\times {10}^{-6}m)$

7. the height of the Mount Everest is $(8.848\times {10}^{3}m)$

NCERT Solutions for Class 8 Maths Chapter 12 Exponents And Powers - Exercise: 12.2

Question: 1(i) Express the following numbers in standard form .

0.0000000000085

Answer:

The standard form is $8.5\times {10}^{-12}$

Question: 1(ii) Express the following numbers in standard form.

0.00000000000942

Answer:

The standard form is $9.42\times {10}^{-12}$

Question: 1(iii) Express the following numbers in standard form.

6020000000000000

Answer:

The standard form is $6.02\times {10}^{15}$

Question: 1(iv) Express the following numbers in standard form.

0.00000000837

Answer:

The standard form of the given number is $8.37\times {10}^{-9}$

Question: 1(v) Express the following numbers in standard form.

31860000000

Answer:

The standard form is $3.186\times {10}^{10}$

Question: 2(i) Express the following numbers in usual form.

$3.02\times {10}^{-6}$

Answer:

$3.02\times {10}^{-6}$

$=\frac{3.02}{1000000}$

$=0.00000302$

this is the usual form

Question: 2(ii) Express the following numbers in usual form.

$4.5\times {10}^4$

Answer:

$4.5\times {10}^4=4.5\times 10000$

$=45000$

this is the usual form

Question:2(iii) Express the following numbers in the usual form.

$3\times {10}^{-8}$

Answer:

$3\times {10}^{-8}$

$\frac{3}{100000000}=0.000000030$

this is the usual form

Question: 2(iv) Express the following numbers in usual form.

$1.0001\times {10}^9$

Answer:

$1.0001\times {10}^9$

$=1.0001\times 100000000$

$=1000100000$

this is the usual form

Question: 2(v) Express the following numbers in usual form.

$5.8\times {10}^{12}$

Answer:

$5.8\times {10}^{12}$

$=5.8\times 100000000000$

$=5800000000000$

this is the usual form

Question:2(vi) Express the following numbers in usual form.

$3.61492\times {10}^6$

Answer:

$3.61492\times {10}^6$

$=3.61492\times 1000000$

$=3614920$

17155

Question:3(i) Express the number appearing in the following statements in standard form.

1 micron is equal to $\frac{1}{1000000}m$ .

Answer:

1 micron is equal to

$\frac{1}{1000000}m$

$=1\times {10}^{-6}$

Question:3(ii) Express the number appearing in the following statements in standard form.

Charge of an electron is 0.000,000,000,000,000,000,16 coulomb.

Answer:

Charge of an electron is 0.000,000,000,000,000,000,16 coulomb.

$=1.6\times {10}^{-19}$ coulomb.

Question:3(iii) Express the number appearing in the following statements in standard form.

Size of a bacteria is 0.0000005 m.

Answer:

Size of a bacteria is 0.0000005 m

$\frac{5}{10000000}=5\times {10}^{-7}m$

Question:3(iv) Express the number appearing in the following statements in standard form.

Size of a plant cell is 0.00001275 m.

Answer:

Size of a plant cell is0.00001275m

$=\frac{1275}{10000}=1.275\times {10}^{-5}m$

Question:3(v) Express the number appearing in the following statements in standard form.

Thickness of a thick paper is 0.07 mm

Answer:

The thickness of a thick paper is 0.07

$=\frac{7}{100}=7\times {10}^{-2}mm$

Question:4 In a stack there are 5 books each of thickness 20mm and 5 paper sheets each of thickness 0.016 mm. What is the total thickness of the stack.

Answer:

the thickness of each book = 20mm

So, the thickness of 5 books = $(5\times 20)=100mm$

the thickness of one paper sheet =0.016mm

So, the thickness of 5 paper sheet = $(5\times 0.016)=0.08mm$

the total thickness of the stack = (100+0.08)mm

=100.08 mm or

$(1.008\times {10}^{-2}mm)$